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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Dec 2007 11:22:41 IST
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Z0=a+b(i) ------(where i root of -1)
find the roots of:
x3-2(1+a)x2 + (4a +a2 +b2)x - 2(a2 +b2)=0
plz give the answer as well as the steps .
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20 Dec 2007 12:47:43 IST
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AREY NO ONE KNOWS THIS ONE...COMMON SOMEONE ANSWER!!!!PLZZZZZZZZZZZZZZZZZZZZZZZZ.....................
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u can easily see that x=2 is a solution of the equation.
so factorizing it , we get
(x-2) ( x 2 - 2ax + a 2 + b 2 ) =0
solving , we get
x=2 , x = a + ib , x= a- ib
or x = 2 , z , z bar
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20 Dec 2007 15:23:41 IST
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nadeem is right .
By hit and trial you get that (x-2) is a factor of the eqn.
Then, you can divide and get,
x = 2, a+ib, a-ib.
i think the question should be z = a+ib,
Then soln. x = 2, z , z bar( conjugate of z)
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